MATH 263 prep sheet

# Dx 4 solve for v then y v 1 1 n 1 most odes having

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Unformatted text preview: mplex Substituting into the Bernoulli equation, we get a linear equation dx for dx v : dv + (1 − n) p( x )v = (1 − n)q( x ) . dx 4. Solve for v , then y = v 1 1− n 1. Most O.D.E’s having the form M ( x, y )dx + N ( x, y ) dy = 0 are not exact. but can be made exact by multiplying by all terms by an integrating factor. Condition Integrating Factor C1e λ1 x + C2 xe λ1 x 1 {eαx cos(βx ),eαx sin(βx)} f ( x ) , assume y p yp Based on f ( x) an x n + an −1 x n −1 + L + a1 x + a0 e αx (a x n n + an −1 x n −1 + L + a1 x + a0 ) eαx cos(βx )(an x n + an −1 x n −1 + L + a1 x + a0 ) +e sin(βx )(bm x + bm −1 x αx Integrating Factors C1e λ1 x + C2e λ2 x C1eα cos(βx ) + C 2eαx sin(βx ) x Undetermined Coefficients v = y1− n , so dv = (1 − n )y − n dy . 3. Homogeneous Solution λ2 x 1 λ = α ± iβ y '+ p ( x) y = q ( x) y n − n dy − n Multiply both sides by y : y + p( x ) y1− n = q( x ) dx + bλ + c = 0 . {e , e } {e λ x , xe λ x } distinct Bernoulli Equations: 2 λ1 x equal Let ay ′′ + by ′ + cy = f ( x ), a, b, c constant Fundamental Solutions λ1 ≠ λ2 2. is found The homogeneous solution is determined by the roots of the characteristic Exact: M ( x, y ) dx + N ( x, y ) dy = 0 is exact if ∂M ( x, y ) = ∂N ( x, y ) . An ∂y ∂x exact solution has an implicit solution g( x, y ) = c , c constant. ⎛ ⎞ d g( x, y ) = ∫ M ( x, y ) dx + ∫ ⎜ N ( x, y ) − ∫ M ( x, y ) dx ⎟dy dy ⎝ ⎠ [ yp by the method of undetermined coefficients or by variation of parameters. dx 1. α is the m m −1 2. Substitute 3. Solve for + L + b1 x + b0...
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## This document was uploaded on 03/10/2014.

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